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Math Help - Predicates and quantifier ..

  1. #1
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    Predicates and quantifier ..

    Can you help me in this question ..

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  2. #2
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  3. #3
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    please if can give me the steps to solve this question ..
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  4. #4
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    Quote Originally Posted by Rozana View Post
    Can you help me in this question ..

    Apparently this is a multiple choice question, judging from your other posts (which make it clearer).
    But still, the wording is a bit puzzling.

    By domain of Q, do mean the extension of Q?
    If so, then what is the domain under discussion? The set of the first three natural numbers, {0,1,2}?

    If so, and we write out the expansion of that existentially quantified formula, it should look like:

    ~Q(0,0,0) V ~Q(0,0,1) V ~Q(0,0,2)

    Then clearly, none of the four will fill the bill.

    But of course, I may be completely misreading the question; it won't be the first time.
    Sorry, I have to go. I can look back later, but maybe before that someone else will help.
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  5. #5
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    Quote Originally Posted by Rozana View Post

    In the question 4

    I think the answer in none and it should be (f,f,t)


    in the statement x=-x is true because 0=(-0)=0


    do you agree with me ??
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  6. #6
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    I need some hints to solve question 3 ..

    and how can I evaluate ~Q(0,0,1)?
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  7. #7
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    Quote Originally Posted by Rozana View Post

    In the question 4

    I think the answer in none and it should be (f,f,t)


    in the statement x=-x is true because 0=(-0)=0


    do you agree with me ??

    sorry the true answer is (t,f,t)


    but another question i don't understand how to solve
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  8. #8
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    Quote Originally Posted by Rozana View Post
    but another question i don't understand how to solve
    The answer to #4 is:

    TFT

    Ax x+1 > x
    is obviously true, where 'x' ranges over real numbers.

    Ax 2x = 3x, is false where 'x' ranges over real numbers, since, e.g. 2*1 does not equal 3*1.

    Ex x = -x, is true where 'x' ranges over real numbers, since 0 = -0, so there is at least one real number x for which x = -x is true.

    /

    The answer to #3 is:

    a.

    You're being asked to translate Ay Q(0 y 0) into a conjunction or disjunction. But the only values allowed for 'y' are 0 or 1.

    So for Q(0 y 0) to hold for all allowed values of 'y' is to say that it holds for both 0 and for 1, when 0 or 1, respectively replace 'y'.

    So, in these circumstances,

    Ay Q(0 y 0)

    becomes

    Q(0 0 0) /\ Q(0 1 0)

    That's all there is to it, really.
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  9. #9
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    Hello Rozana

    Q(x,y,z) is an example of a propositional function; that is, a function that returns a truth value (i.e. the value True or False) that will depend upon the value(s) of any parameter(s) that are supplied to it.

    Here, the value of Q will be determined by the values of three parameters, x, y and z. The values that x, y, z can take are given in the question - defined as the domain of Q(x,y,z).

    Since x can take 3 different values (0,1,2) ; and y can take 2 different values (0,1); and z can also take 2 values (0,1), there are 3\times2\times2= 12 possible propositions that can be represented by Q(x,y,z), each one returning the value True or False. They are:
    Q(0,0,0),\; Q(0,0,1),\; Q(0,1,0),\;Q(0,1,1)

    Q(1,0,0),\; Q(1,0,1),\; Q(1,1,0),\;Q(1,1,1)


    Q(2,0,0),\; Q(2,0,1),\; Q(2,1,0),\;Q(2,1,1)

    Without knowing the actual details of what Q is, we can't say what value each of these propositions has. However, we can say, that if (as in your original question) \exists z,\; \neg Q(0,0,z) is a true statement, then for one or both of the possible values of z ( 0 and 1), \neg Q(0,0,z) is true. In other words, \neg Q(0,0,0) is true or \neg Q(0,0,1) is true (or both). So:
    \exists z,\; \neg Q(0,0,z) can be written as \neg Q(0,0,0) \lor \neg Q(0,0,1)
    ****

    Question 3 can be answered in the same way:
    \forall y,\;Q(0,y,0)
    means
    For all y (that is, y = 0 and y = 1), Q(0,y,0) is true.
    So
    Q(0,0,0) is true and Q(0,1,0) is true.
    Therefore:
    \forall y,\;Q(0,y,0) can be written as Q(0,0,0) \land Q(0,1,0)
    ****

    Question 4


    Examine the truth of each of the propositions in turn:
    p:\forall x, \; x+1>x
    p is True. It simply says that if we add 1 to any real number, the answer is greater than the number we started with.
    q:\forall x, \;2x = 3x
    Clearly q is False. To show this, we only need find one value of x for which 2x \ne 3x. (That's not very hard, is it?)
    r:\exists x,\;x=-x
    r is True. There is a value of x for which x = -x; that is x = 0.

    So the ordered triple (p,q,r) has the truth value (T, F, T).


    Does that clear up the problems?


    Grandad
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  10. #10
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    Thank you for all especially Grandad..

    very clear explanation ..
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